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I've been trying to understand how a differential amplifier works. In this article https://electronicspost.com/what-is-a-differential-amplifier-discuss-the-operation-of-a-differential-amplifier/ it says that from the perspective of Vin1, transistor Q1 acts as a common emitter amplifier and a common collector amplifier, and Q2 acts as a common base amplifier. I don't see how this can be the case. My understanding is that in a common emitter amplifier, the emitter is connected to the base and the collector. Where is the connection in the differential amplifier? Same goes for the transistor acting as common base or common collector. In a common collector, shouldn't the collector be connected to the base and emitter? Where is the connection? In what sense is the transistor acting as a common collector?

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  • \$\begingroup\$ in a common emitter amplifier, the emitter is connected to the base and the collector huh? The emitter is common to input (the other "input terminal" being base) and output (the other "output terminal" being collector). \$\endgroup\$
    – greybeard
    Commented Dec 17, 2022 at 10:44
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    \$\begingroup\$ It doesn't say either of the things you said the article contained. If you are going to quote, do it verbatim. Total accuracy is required. Even better, make a screen shot of the words in the article and post that image. Don't paraphrase because it becomes impossible to tell where in the article it came from and, leads to inaccuracies that wastes time. \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2022 at 10:49
  • \$\begingroup\$ You did not change the part about basic BJT amplifier configurations/"topologies". It seems you would profit from revisiting them, using a simulator or the real thing for some tinkering. \$\endgroup\$
    – greybeard
    Commented Dec 17, 2022 at 10:59
  • \$\begingroup\$ @greybeard I don't understand what you mean. What part of BJT amplifiers have I misunderstood? Am I wrong that, for example, in a common emitter the emitter is connected to the collector and base? \$\endgroup\$
    – Urthona26
    Commented Dec 17, 2022 at 11:02
  • \$\begingroup\$ Yes, I see the part about the common terminals being "connected" to both others as a mis-presentation. It may be just language: You can use the schematic tool from the post editor tool bar to add a schematic. (You can even do basic simulation - but check out circuitlab.com.) \$\endgroup\$
    – greybeard
    Commented Dec 17, 2022 at 11:15

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The answer to your question depends on the ouput node which is taken into consideation:

  • Yes - it is true that "from the perspective of Vin1, transistor Q1 acts as a common emitter amplifier" - when the output is taken from the collector of Q1. In this case, the input resistance of Q2 acts as a emitter resistance for Q1 causing negative feedback (and reduces the gain by a factor of 0.5)-

  • However, if the output signal is taken from the collector of Q2 (input still at the base of Q1) , we have a two-transistor combination. The input for Q2 is the emitter node which is connected to the emitter of Q1. Hence, we have a series combination of two transistors - Q1 working as a common collector stage (output emitter) and Q2 works in common base operation (input emitter).

  • Because of symmetric properties of the circuit, we can exchange the operation Q1 and Q2.

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  • \$\begingroup\$ Which resistor are you referring to when you say 'the input resistance of Q2 acts as an emitter resistance for Q1'? R_E? \$\endgroup\$
    – Urthona26
    Commented Dec 17, 2022 at 11:07
  • \$\begingroup\$ Of course, the input resistance at the emitter of Q2, which is rather small. It is 1/gm=Vt,/Ic \$\endgroup\$
    – LvW
    Commented Dec 17, 2022 at 16:59

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